A and B are two weak students of the course “Probability Theory” and their chances of solving a problem in “Probability Theory” correctly are 1/6 and 1/8, respectively. If the probability of their making a common error is 1/525 and they obtain the same answer, find the probability that their answer is correct.
I don't understand how to proceed with this question. Is the probability of common error relevant? Any solutions/tips welcome. Thank you.
Let $A$ be the event that person A got the question right, and $B$ the event that person B got the question right.
Let $C$ be the event that person $A$ and $B$ got the same answer.
We are told the following values in the question:
$Pr(A)=\frac{1}{6},~Pr(B)=\frac{1}{8},~Pr(A^c\cap B^c\cap C)=\frac{1}{525}$
It is heavily implied from context that $Pr(C\mid A\cap B) = 1$ and that $Pr(C\mid A\cap B^c)=Pr(C\mid A^c\cap B)=0$. After all, if $A$ and $B$ both got the right answer, we can assume that the answer was the same (invalid assumption in the case of questions have multiple different right answers). Further, if $A$ got a right answer and $B$ did not, then we can assume that their answers were different, and similarly so if $B$ had the right answer while $A$ didn't.
We are tasked with calculating $Pr(A\cap B\mid C)$
To continue, we might try to approach directly via definitions
$$Pr(A\cap B\mid C) = \dfrac{Pr(A\cap B\cap C)}{Pr(C)}$$
Noting everything from above, this can simplify to
$$\dfrac{Pr(A\cap B)}{Pr(A\cap B)+Pr(A^c\cap B^c\cap C)}$$
Unfortunately, there doesn't appear to be any way to calculate $Pr(A\cap B)$ from the given information however so we are stuck here. If we were to assume that $A$ and $B$'s correctness is independent of one another, then we could expand $Pr(A\cap B)$ as $Pr(A)\times Pr(B)$ and continue, but this is a rather strong assumption that the problem doesn't imply that we should make.
If we were to continue anyways with that assumption, then this continues as:
$$\dfrac{Pr(A)\times Pr(B)}{Pr(A)\times Pr(B)+Pr(A^c\cap B^c\cap C)}=\dfrac{\frac{1}{6}\cdot \frac{1}{8}}{\frac{1}{6}\cdot \frac{1}{8}+\frac{1}{525}}$$ which can then be simplified down to a single value.